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JEE MAIN - Physics (2018 - 15th April Evening Slot - No. 11)

A thin uniform bar of length $$L$$ and mass $$8$$ m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2$$\upsilon $$ and $$\upsilon $$ respectively. The masses stick to the bar after collision at a distance $${L \over 3}$$ and $${L \over 6}$$ respectively from the center of the bar. If the br starts rotating about its center of mass as a result of collision, the angular speed of the bar will be :
JEE Main 2018 (Online) 15th April Evening Slot Physics - Rotational Motion Question 202 English
$${v \over {5L}}$$
$${6v \over {5L}}$$
$${3v \over {5L}}$$
$${v \over {6L}}$$

Explanation

Centre of mass after the collision will be at O. Therefore,

$$2m{L \over 6} = m{L \over 3}$$

Now, using conservation of angular momentum, we get

$$2m{L \over 6} \times v + m{L \over 3} \times 2v = I\omega $$

Therefore,

$$2m{L \over 6} \times v + m{L \over 3} \times 2v = \left[ {8m{{{L^2}} \over {12}} + 2m{{\left( {{L \over 6}} \right)}^2} + m{{\left( {{L \over 3}} \right)}^2}} \right]\omega $$

$$ \Rightarrow \left( {{L \over 6} + {L \over 3}} \right)2mv = \left[ {8m{{{L^2}} \over {12}} + {{2m{L^2}} \over {36}} + {{m{L^2}} \over 9}} \right]\omega $$

$$ \Rightarrow {{3L} \over 6} \times 2mV = {5 \over 6}m{L^2}\omega $$

$$ \Rightarrow Lmv = {5 \over 6}m{L^2}\omega \Rightarrow \omega = {{6Lmv} \over {5m{L^2}}} \Rightarrow \omega = {6 \over 5}{v \over L}$$

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